$\begingroup$I'm still trying to verify. My idea was to construct a sufficiently path disconnected continuum while ensuring that every path between two particular points must avoid a fat Cantor set (in fact, so that $\gamma\cap P$ has measure $0$). With the Pseudoarc you may be able to construct paths $\gamma$ such that $\gamma \cap P$ has large measure.$\endgroup$
– D.S. LiphamAug 4 '18 at 21:45

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$\begingroup$I think that you will want to approximate by finite unions of Cantor Sets producted with $I$. Then show that near each point that has arc-free neighborhoods the continuum can be approximated by some sufficiently hefty collection of these sets so that you can use polygonal approximations of your arc 'going against the grain' enough to get fat in the limit. There may also be a slick argument using the Sierpinski Carpet. It is an attractive question. You may like: Can you draw a path between two points in the complement of planar Brownian motion crossing the trace only finitely many times?$\endgroup$
– John SamplesAug 5 '18 at 6:24

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$\begingroup$@JohnSamples I do not understand your comment --- it seems that in your example there is an everywhere dense curve (?) --- if yes then the condition holds.$\endgroup$
– Anton PetruninAug 5 '18 at 15:53

$\begingroup$I am not sure about whether your conjecture is true, but it is the proof method that seems most promising to me. I don't know any results that it follows cleanly from, though there may be one. The Brownian Motion problem is totally unrelated, just thought you would like it. To my knowledge it is an open problem.$\endgroup$
– John SamplesAug 6 '18 at 20:19

$\begingroup$@AntonPetrunin Is there any progress in higher dimensions than 2? It seems I can use it in my research. This result helps me understand the structure of a certain quotient space that I construct.$\endgroup$
– Piotr HajlaszDec 11 '19 at 22:36

1 Answer
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The answer to this question is positive. A required path $\gamma$ can be constructed inductively using the following

Lemma. For any continuum $P\subset\mathbb R^2$, distinct points $x,y\in P$, and $\varepsilon>0$ there are continua $P_1,\dots,P_{k}\subset P$ of diameter $<\varepsilon$, and horizontal or vertical arcs $A_0,\dots,A_{k+1}$ in the plane such that

1) $\{x\}=A_0$, $\{y\}=A_{k+1}$;

2) $\sum_{i=1}^k\lambda(A_i\setminus P)<\varepsilon$;

3) for every $i\in\{1,\dots,k\}$ the continuum $P_{i}$ intersects the arcs $A_{i-1}$ and $A_i$.

Proof. Replacing $P$ by a smaller subcontinuum, we can assume that $P$ is irreducible between points $x$ and $y$, which means that any subcontinuum of $P$ that contains $x$ and $y$ coincides with $P$.

The irreducibility of $P$ implies that for any rational numbers $a<b$ the (closed) set $\{x\in \mathbb R:\{x\}\times[a,b]\subset P\}$ is nowhere dense in $\mathbb R$. Consequently the set $$G_1:=\{x\in\mathbb R:P\cap(\{x\}\times\mathbb R)\mbox{ is zero-dimensional}\}$$is dense $G_\delta$ in $\mathbb R$. Choose an increasing sequence $a_0,\dots,a_n\in G_1$ such that $P\subset [a_0,a_n]\times\mathbb R$ and $|a_i-a_{i-1}|<\frac12\varepsilon$ for all positive $i\le n$. For every $i\in\{0,\dots,n\}$ the inclusion $a_i\in G_1$ implies that the set $Z_i:=\{y\in \mathbb R:(a_i,y)\in P\}$ is zero-dimensional and hence nowhere dense in $\mathbb R$.

By analogy, the set $$G_2:=\{y\in\mathbb R:P\cap(\mathbb R\times \{y\})\mbox{ is zero-dimensional}\}$$is dense $G_\delta$ in $\mathbb R$. Then we can choose a monotone sequence $b_0,\dots,b_m\in\mathbb G_2\setminus\bigcup_{i=0}^nZ_i$ such that $P\subset \mathbb R\times [b_0,b_m]$ and $|b_i-b_{i-1}|<\frac12\varepsilon$ for all positive $i\le m$.

For every $(i,j)\in\{1,\dots,n\}\times\{1,\dots,m\}$ consider the cube $Q_{i,j}:=[a_{i-1},a_i]\times[b_{j-1},b_j]$ and its boundary $\partial Q_{i,j}$ in the plane. The choice of the points $a_{i-1},a_i,b_{j-1},b_j$ guarantees that the intersection $P\cap\partial Q_{i,j}$ is zero-dimensional and does not contain the vertices of the cube $Q_{i,j}$.

Let $V:=\bigcup_{i=0}^n\{a_i\}\times [b_0,b_m]$ and $H:=\bigcup_{j=0}^m[a_0,a_n]\times\{b_j\}$ be the unions of vertical and horizontal sides of the cubes $Q_{i,j}$. Let $\mathcal C$ be the family of connected components of the symmteric difference $\Xi:=(V\cup H)\setminus (V\cap H)$. Each component $C\in\mathcal C$ is a vertical or horizontal interval of length $<\frac12\varepsilon$. By the regularity of the Lebesque measure on the real line, there is a disjoint finite family $\mathcal I$ of compact connected subsets of $\Xi\cup\{x,y\}$ such that

$\bullet$ $\{x\},\{y\}\in\mathcal I$;

$\bullet$ $P\cap\Xi\subset \bigcup\mathcal I$;

$\bullet$ $\lambda(\bigcup\mathcal I\setminus P)<\varepsilon$;

$\bullet$ each set $I\in\mathcal I$ meets the set $P$.

Now consider the finite graph $\Gamma$ whose set of vertices coincides with $\mathcal I$ and two distinct intervals $I,J\in\mathcal J$ form an edge of the graph $\Gamma$ if there exists a square $Q_{i,j}$ and a connected component of $P\cap Q_{i,j}$ that intersects both arcs $I$ and $J$. It can be shown that the graph $\Gamma$ is connected, so we can find a sequence $\{x\}=A_0,A_1,\dots,A_{k+1}=\{y\}$ of pairwise distinct vertices of the graph $\Gamma$ such that for every $i\le k$ the unordered pair $\{A_i,A_{i+1}\}$ is an edge of $\Gamma$. The latter means that $A_{i-1}$ and $A_i$ are intersected by some subcontinuum $P_i$ of $P$ that is contained in some square $Q_{j,l}$ and hence has diameter $<\varepsilon$.

It is clear that the sequences $P_1,\dots,P_k$ and $A_0,\dots,A_{k+1}$ have the required properties.

Remark. Unfortunately, the proof of the Lemma essentially uses the planarity of the continuum $P$. It would be interesting to know if the problem of Anton Petrunin still has affirmative answer for continua in arbitrary (finite-dimensional) Banach spaces.

$\begingroup$I am confused. If $x = (0,0)$, $y = (1,1)$, and $P$ is the line segment with endpoints $x$ and $y$, then it seems that the conclusion of your lemma is impossible. Am I missing something, or is there a problem with the lemma?$\endgroup$
– Will BrianAug 13 '18 at 16:13

$\begingroup$@WillBrian Why impossible? Just divide the line segment $P$ into consequtive subsegments $P_1,\dots,P_{k}$ of smaller length and at junction points attach short vertical or horizontal arcs $A_1,\dots,A_{k-1}$. What is the problem? In fact, those arcs $A_1,\dots,A_{k-1}$ can be degenerated (i.e., singletons).$\endgroup$
– Taras BanakhAug 13 '18 at 16:22

$\begingroup$Thanks for explaining. I was thinking that $A_i$ had to be a subset of $P_i$. (You don't say this, so it was my fault for not reading carefully enough.)$\endgroup$
– Will BrianAug 13 '18 at 16:28

$\begingroup$@WillBrian In fact, those arcs $A_i$ form a part of the path $\gamma$ connecting the points $x$ and $y$. So, $\gamma$ consists of countably many horizontal or vertical arcs $A_i$ plus some Cantor set in $P$, whose points are intersections of decreasing sequences of continua $P_{i_1}\supset P_{i_1i_2}\supset P_{i_1i_2i_3}\supset$, constructed inductively with help of trees. The inductive construction is clear but is rather long. So, at the moment, I have written only the proof of Lemma, hoping that the interested reader can complete the inductive construction of $\gamma$ (based on this lemma).$\endgroup$
– Taras BanakhAug 13 '18 at 16:37

$\begingroup$@Anton-Petrunin Thank you for +550 bounty! Are you planning to use this result in your research?$\endgroup$
– Taras BanakhAug 15 '18 at 4:39